Can a local variable's memory be accessed outside its scope?

user_548360 0
Viewed 293437

I have the following code.

#include <iostream>

int * foo()
{
    int a = 5;
    return &a;
}

int main()
{
    int* p = foo();
    std::cout << *p;
    *p = 8;
    std::cout << *p;
}

And the code is just running with no runtime exceptions!

The output was 58

How can it be? Isn't the memory of a local variable inaccessible outside its function?

dangling-pointer local-variables memory-management c++
20 Answers

Your code is very risky. You are creating a local variable (wich is considered destroyed after function ends) and you return the address of memory of that variable after it is destoyed.

That means the memory address could be valid or not, and your code will be vulnerable to possible memory address issues (for example segmentation fault).

This means that you are doing a very bad thing, becouse you are passing a memory address to a pointer wich is not trustable at all.

Consider this example, instead, and test it:

int * foo()
{
   int *x = new int;
   *x = 5;
   return x;
}

int main()
{
    int* p = foo();
    std::cout << *p << "\n"; //better to put a new-line in the output, IMO
    *p = 8;
    std::cout << *p;
    delete p;
    return 0;
}

Unlike your example, with this example you are:

  • allocating memory for int into a local function
  • that memory address is still valid also when function expires, (it is not deleted by anyone)
  • the memory address is trustable (that memory block is not considered free, so it will be not overridden until it is deleted)
  • the memory address should be deleted when not used. (see the delete at the end of the program)
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